After one minute there is \(y\) gallons in the tank 2, so the capacity of the tank 2 is \(2(x-y)\) gallons.

Obviously, each minute the tank 1 is filled with \(x-y\) gallons of water, and tank 2 is filled with \(y\) gallons.

Let A is the number of minutes after which the 1 tank is full, and B is the number of minutes after which 2 the 2 tank is full.

Then:

\(A(x-y)=z\)

\(By=2(x-y)\)

\(A=\frac{z}{x-y}\)

\(B=\frac{2(x-y)}{y}\)

We need to compare A and B, so we are comparing \(\frac{z}{x-y}\) and \(\frac{2(x-y)}{y}\)

\(\frac{z}{x-y}\) ... \(\frac{2(x-y)}{y}\)

\(yz\) ... \(2(x-y)(x-y)\)

\(yz\) ... \(2x^2-4xy+2y^2\)

If (1) is true, then \(yz< 2x^2-4xy-y^2\)

Since \(2x^2-4xy+2y^2> 2x^2-4xy-y^2\), then \(yz <2x^2-4xy+2y^2\) and we are able to compare two time periods. The statement (1) alone is susfficient.

If (2) is true, then \(2(x-y)<0.5z\)

\(\frac{z}{x-y}>4\)

This means that \(A>4\)

However, \(B=\frac{2(x-y)}{y}\), so there is no z and we only know that \(x>y\), but nothing could be said to compare \(x-y\) and \(y\). For example, if \(x=2y\), then \(B=2\) and \(A>B\). However, if \(x=5y\), then \(B=8\) and we could not compare A and B.

So, the answer is (A)

NOTE: You should post DS problems in the other forum.

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